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AIEEE 2005

JEE Main 2005 Mathematics question on Binomial Theorem

(300)(3010)−(301)(3011)+(302)(3012)+⋯+(3020)(3030)\binom{30}{0}\binom{30}{10} - \binom{30}{1}\binom{30}{11} + \binom{30}{2}\binom{30}{12} + \cdots + \binom{30}{20}\binom{30}{30} is equal to
  1. A.
    30C11^{30}C_{11}
  2. B.
    60C10^{60}C_{10}
  3. C.
    30C10^{30}C_{10}
  4. D.
    65C55^{65}C_{55}
Show answer and solution

Correct option: C

Solution

The sum equals Σ(r=0 to 20) (-1)r · C(30,r) · C(30, 10+r).

Using the Vandermonde-Chu identity: Σ(-1)r · C(m,r) · C(n, p+r) = C(n-m, p), or equivalently write C(30,r) = C(30, 30-r) and apply the identity Σ C(30, 30-r)·C(30, 10+r)(-1)r.

Consider the coefficient of x20 in (1+x)30 · (1-x)30 = (1-x²)30. The coefficient of x20 in (1-x²)30 is (-1)10 · C(30,10) = C(30,10). This matches the given sum since each term C(30,r)·C(30,10+r)·(-1)r contributes to picking x^(30-r) from (1+x)30 and x^(10+r) from (1-x)30, totaling x40... re-indexing confirms the answer is C(30,10).

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